91Ó°ÊÓ

The moment of inertia is a crucial concept when dealing with rotational dynamics, much like mass is for linear motion. It represents how difficult it is to change the rotational state of an object. For the given system, consisting of a rod and hoop, we need to calculate the moment of inertia around the axis through the center of the hoop.
The moment of inertia (\(I\)) of the rod about its center is \(I_{\text{rod}} = \frac{1}{3}ml^2\). However, because the point of rotation shifts due to the rolling hoop, we utilize the parallel axis theorem. This theorem states that \(I = I_{\text{center}} + m d^2\), where \(d\) is the distance from the center of mass to the new axis.
In our problem, we find \(d\) as the distance from the rod's center to the rotational axis, adding \(ml^2\) to \(I_{\text{rod}}\). Thus, the complete moment of inertia for this scenario is \(I = \frac{1}{3}ml^2 + ml^2 = \frac{4}{3}ml^2\).
This inertia is vital for calculating the system's oscillatory behavior as it influences how fast it oscillates when displaced from equilibrium.

Simple harmonic motion (SHM) describes a type of oscillatory motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. In this context, it is important because the period of oscillation we're calculating is a result of SHM.
When the rod attached to the hoop is slightly displaced, gravitational torque acts to restore it to the equilibrium position. The equation \[-\tau = I \alpha\] encapsulates this where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration.
The torque due to gravity when displaced by a small angle \(\theta\) is \(mgl \theta\) (approximating \(\sin \theta \approx \theta\) for small angles).
This results in the differential equation for angular SHM: \[\frac{d^2\theta}{dt^2} + \left(\frac{3g}{4l}\right)\theta = 0\].
This represents SHM because the second derivative of angle \(\theta\) (angular acceleration) is proportional and opposite to \(\theta\) itself, indicating oscillation around the equilibrium point.

Angular motion involves the rotation of an object around a specific axis and is key to understanding the dynamics of our rod and hoop system. In our exercise, the hoop rolls without slipping, making angular motion central to determining the oscillation period.
The relationship underpinning this scenario is Newton's second law for rotation, which states \(\tau = I \alpha\), linking torque \(\tau\), the moment of inertia \(I\), and angular acceleration \(\alpha\).
Here, \(\alpha\), the angular acceleration, is the change accelerated rotation due to the gravity-induced torque. We substitute this into our earlier derivation to find \(\alpha = -\frac{3g}{4l} \theta\).
From SHM's characteristics, the angular frequency \(\omega\) is \(\sqrt{\frac{3g}{4l}}\) and directly influences the oscillation period \(\tau\). Calculating the period involves finding \(\tau = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4l}{3g}}\).
Understanding angular motion in this context helps explain why the system has a specific period of oscillations, combining both linear and rotational dynamics principles seamlessly.